1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)

=>4x+8=3x-1

=>x=-9

2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)

=>8x-4=5x-7

=>3x=-3

=>x=-1

3: ĐKXD: x>=0

\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

=>\(x+\sqrt{x}-6=x-1\)

=>căn x=-1+6=5

=>x=25

4: ĐKXĐ: x>=0

PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

=>x-2*căn x-3=x-4

=>-2căn x-3=-4

=>2căn x+3=4

=>2căn x=1

=>căn x=1/2

=>x=1/4

1: ĐKXĐ: \(-1< x< 1\)

2: ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)

3: ĐKXĐ: \(\left[{}\begin{matrix}x< -3\\x\ge2\end{matrix}\right.\)

4: ĐKXĐ: \(2< a\le3\)

\(A=P:Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}:\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}=1+\dfrac{-5}{\sqrt{x}+4}\)

Điều kiện : \(x\ge4\Rightarrow\sqrt{x}+4\ge4\Rightarrow-\dfrac{5}{\sqrt{x}+4}\le-\dfrac{5}{4}\Rightarrow\dfrac{5}{\sqrt{x}+4}\ge\dfrac{5}{4}\)

Dấu ''='' xảy ra \(\Leftrightarrow x=0\)

Vậy \(min_A=\dfrac{5}{4}\Leftrightarrow x=0\)

Điều kiện: x>2

P= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{2}+2}{\sqrt{x}-1}\right)\)

P= \(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b) P= \(\dfrac{1}{4}\)

⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)

⇔\(4\sqrt{x}-8=3\sqrt{x}\)

⇔\(\sqrt{x}=8\)

⇔x=64 (TM) 

Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)

c??

Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{x-4}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)

\(=\dfrac{-1}{\sqrt{x}-2}\)

Để P nguyên thì \(\sqrt{x}-2\in\left\{-1;1\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3\right\}\)

hay \(x\in\left\{1;9\right\}\)

Mik ko hiểu đề

ĐKXĐ: x>=0; x<>1

a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)

\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)

\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)

c: B=2/3

=>căn x/căn x+1=2/3

=>căn x=2

=>x=4

d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)

=>B<1

e: B>1

=>-1/căn x+1>0

=>căn x+1<0(vô lý)

=>KO có x thỏa mãn

f: B nguyên khi căn x chia hết cho căn x+1

=>căn x+1-1 chia hết cho căn x+1

=>căn x+1=1 hoặc căn x+1=-1(loại)

=>căn x=0

=>x=0

\(a,P\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\\ =\dfrac{1}{\sqrt{x}}.\dfrac{\sqrt{x}-2}{3}\\ =\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

\(c,P=\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4\left(\sqrt{x}-2\right)-3\sqrt{x}}{12\sqrt{x}}=0\\ \Leftrightarrow4\sqrt{x}-8-3\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}=8\\ \Leftrightarrow x=64\left(tmdk\right)\)

Vậy \(x=64\) thì \(P=\dfrac{1}{4}\)

a: Ta có: \(2\sqrt{2}-\dfrac{1}{2}\cdot\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\cdot\dfrac{1}{2}=2\sqrt{2}\)

\(\Leftrightarrow\sqrt{x}=4\sqrt{2}\)

hay x=32

b: Ta có: \(2\sqrt{x}-\sqrt{\dfrac{x}{3}}=1\)

\(\Leftrightarrow2\sqrt{x}-\dfrac{\sqrt{3}}{3}\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\dfrac{6+\sqrt{3}}{11}\)

hay \(x=\dfrac{39+12\sqrt{3}}{121}\)

c: Ta có: \(4\sqrt{x}+\sqrt{\dfrac{x}{2}}=\dfrac{1}{3}\)

\(\Leftrightarrow4\sqrt{x}+\dfrac{\sqrt{2}}{2}\sqrt{x}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{x}=\dfrac{8-\sqrt{2}}{93}\)

hay \(x=\dfrac{66-16\sqrt{2}}{8649}\)

\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)

\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)

Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)

a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)

=>x+6<0

=>x<-6

b: ĐKXĐ: (-2)/(6-x)>=0

=>6-x<0

=>x>6

c: ĐKXĐ: (-x+3)/(-6)>=0

=>-x+3<=0

=>-x<=-3

=>x>=3

d: ĐKXĐ: (7x-1)/-9>=0

=>7x-1<=0

=>x<=1/7

e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0

=>x+2>=0

=>x>=-1

f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0

=>x-2>=0

=>x>=2